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By

Dr. Attaullah Shah

Swedish College of Engineering and TechnologyWah Cantt.

CE-401

Reinforced Concrete Design-II

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Course Outline:

Analysis & design of axially loaded columns, Eccentrically

loaded columns by USD

Analysis & design of strip footing for wall, spread footings

for columns and combined footings by USD.

Design of retaining wall. Introduction to limit states.

Detailing of reinforcement.

Introduction to design of staircases and water tanks.

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Columns subjected to eccentric loadings

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Eccentric Compression

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Interaction diagrams of combined bending and compression

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Behavior under Combined Bending

and Axial Loads

Interaction Diagram Between Axial Load and Moment

( Failure Envelope )

Concrete crushes

before steel yields

Steel yields before concrete

crushes

Any combination of P and M outside the envelope will cause failure.

Note:

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Behavior under Combined Bending and

Axial Loads

Axial Load and Moment Interaction DiagramGeneral

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Behavior under Combined

Bend ing and Axial LoadsResultant Forces action at Centroid

( h/2 in this case )

s2

positiveisncompressiocs1n

TCCP

Moment about geometric center

2*

22*

2* 2s2c1s1n

hdT

ahCd

hCM

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Columns in Pure Tens ion

Section is completely cracked (no concrete

axial capacity)

Uniform Strainy

N

1ii

sytensionn AfP

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Columns

Strength Reduction Factor, f (ACI Code 9.3.2)

Axial tension, and axial tension with flexure.

f = 0.9

Axial compression and axial compression with

flexure.

Members with spiral reinforcement confirmingto 10.9.3 f 0.70

Other reinforced members f 0.65

(a)

(b)

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Columns

Except for low values of axial compression, f may be

increased as follows:

when and reinforcement is symmetric

and

ds = distance from extreme tension fiber to centroid of

tension reinforcement.

Then f may be increased linearly to 0.9 as fPn

decreases from 0.10fc Ag to zero.

psi000,60y f

70.0

s

h

ddh

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Column

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Columns

Commentary:

Other sections:

f may be increased linearly to 0.9 as thestrain s increase in the tension steel. fPb

D i f C bi d B d i d

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Design fo r Combined Bend ing and

Axial Load (Sho rt Column)

Design - select cross-section and reinforcement

to resist axial load and moment.

D i f C bi d B d i d

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Design fo r Combined Bend ing and

Axial Load (Sho rt Column)

Column Types

Spiral Column - more efficient for e/h < 0.1,

but forming and spiral expensive

Tied Column - Bars in four faces used when

e/h < 0.2 and for biaxial bending

1)

2)

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General Procedure

The interaction diagram for a column is

constructed using a series of values for Pn and

Mn. The plot shows the outside envelope of theproblem.

G l P d f

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General Procedure fo r

Cons truct ion of ID

Compute P0 and determine maximum Pn in compression

Select a c value (multiple values)

Calculate the stress in the steel components.

Calculate the forces in the steel andconcrete,Cc, Cs1 and Ts.

Determine Pn value.

Compute the Mn about the center.

Compute moment arm,e = Mn / Pn.

G l P d f

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General Procedure fo r

Construct ion of ID

Repeat with series of c values (10) to obtain a series of values.

Obtain the maximum tension value.

Plot Pn verse Mn.

Determine fPn and fMn.

Find the maximum compression level.

Find the f will vary linearly from 0.65 to 0.9for the strain values

The tension component will be f = 0.9

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Example: Ax ial Load vs . Moment

In terac t ion DiagramConsider an square column (20 in x 20 in.) with 8 #10

(r = 0.0254) and fc = 4 ksi and fy = 60 ksi. Draw the

interaction diagram.

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In teract ion DiagramGiven 8 # 10 (1.27 in2) and fc = 4 ksi and fy = 60 ksi

2 2st

2 2

g

2

st

2

g

8 1.27 in 10.16 in

20 in. 400 in

10.16 in 0.0254400 in

A

A

AA

r

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In teract ion DiagramGiven 8 # 10 (1.27 in2) and fc = 4 ksi and fy = 60 ksi

0 c g st y st

2 2

2

0.85

0.85 4 ksi 400 in 10.16 in

60 ksi 10.16 in

1935 k

P f A A f A

n 0

0.8 1935 k 1548 k

P rP

[ Point 1 ]

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In teract ion DiagramDetermine where the balance point, cb.

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In teract ion DiagramDetermine where the balance point, cb. Using similartriangles, where d = 20 in.2.5 in. = 17.5 in., one can

find cbb

b

b

17.5 in.0.003 0.003 0.00207

0.00317.5 in.

0.003 0.0020710.36 in.

c

c

c

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In teract ion DiagramDetermine the strain of the steel

bs1 cu

b

bs2 cu

b

2.5 in. 10.36 in. 2.5 in.

0.00310.36 in.

0.00228

10 in. 10.36 in. 10 in. 0.00310.36 in.

0.000104

c

c

cc

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In teract ion DiagramDetermine the stress in the steel

s1 s s1

s2 s s1

29000 ksi 0.00228

66 ksi 60 ksi compression

29000 ksi 0.000104

3.02 ksi compression

f E

f E

Example: Axial Load vs Moment

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Example: Axial Load vs . Moment

In terac t ion DiagramCompute the forces in the column

c c 1

s1 s1 s1 c

2

2

s2

0.85

0.85 4 ksi 20 in. 0.85 10.36 in.

598.8 k

0.85

3 1.27 in 60 ksi 0.85 4 ksi

215.6 k

2 1.27 in 3.02 ksi 0.85 4 ksi

0.97 k neglect

C f b c

C A f f

C

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In teract ion DiagramCompute the forces in the column

2s s s

n c s1 s2 s

3 1.27 in 60 ksi

228.6 k

599.8 k 215.6 k 228.6 k 585.8 k

T A f

P C C C T

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Example: Ax ial Load vs. Moment

In terac t ion Diagram

Compute the moment about the center

c s1 1 s 32 2 2 2

0.85 10.85 in.20 in.599.8 k2 2

20 in.215.6 k 2.5 in.

220 in.

228.6 k 17.5 in.2

6682.2 k-in 556.9 k-ft

h a h hM C C d T d


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A single point from interaction diagram,(585.6 k, 556.9 k-ft). The eccentricity of the point is

defined as

6682.2 k-in11.41 in.

585.8 k

Me

P

[ Point 2 ]

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In teract ion DiagramNow select a series of additional points by selectingvalues of c. Select c = 17.5 in. Determine the strain

of the steel. (c is at the location of the tension steel)

s1 cu

s1

s2 cu

s2

2.5 in. 17.5 in. 2.5 in.0.003

17.5 in.

0.00257 74.5 ksi 60 ksi (compression)

10 in. 17.5 in. 10 in.0.003

17.5 in.

0.00129 37.3 ksi (compression)

c

c

f

c

c

f

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c c 1

2

s1 s1 s1 c

2

s2

0.85 0.85 4 ksi 20 in. 0.85 17.5 in.

1012 k

0.85 3 1.27 in 60 ksi 0.85 4 ksi

216 k

2 1.27 in 37.3 ksi 0.85 4 ksi

86 k

C f b c

C A f f

C

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2

s s s

n

3 1.27 in 0 ksi

0 k

1012 k 216 k 86 k

1314 k

T A f

P


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c s1 12 2 2

0.85 17.5 in.20 in.1012 k

2 2

20 in.

216 k 2.5 in.2

4213 k-in 351.1 k-ft

h a hM C C d


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A single point from interaction diagram,

(1314 k, 351.1 k-ft). The eccentricity of the point is

defined as

4213 k-in3.2 in.

1314 k

Me

P

[ Point 3 ]

Example: Ax ial Load vs Moment

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Example: Ax ial Load vs. Moment


Select c = 6 in. Determine the strain of the steel, c =6 in.

s1 cu

s1

s2 cu

s2

s3 cu

2.5 in. 6 in. 2.5 in.0.003

6 in.

0.00175 50.75 ksi (compression)

10 in. 6 in. 10 in.0.003

6 in.

0.002 58 ksi (tension)

17.5 in. 6 in.

c

c

f

c

c

f

c

c

s3

17.5 in.0.003

6 in.

0.00575 60 ksi (tension)f

Example: Ax ial Load vs Moment

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c c 1

s1 s1 s1 c

2

2

s2

0.85

0.85 4 ksi 20 in. 0.85 6 in.

346.8 k

0.85

3 1.27 in 50.75 ksi 0.85 4 ksi

180.4 k C2 1.27 in 58 ksi

147.3 k T

C f b c

C A f f

C

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Example: Axial Load vs. Moment

In teract ion Diagram


c s1 1 s 32 2 2 2

0.85 6 in.346.8 k 10 in.

2

180.4 k 10 in. 2.5 in.

228.6 k 17.5 in. 10 in.

5651 k-in 470.9 k-ft

h a h hM C C d T d

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Example: Axial Load Vs. Moment


A single point from interaction diagram,

(151 k, 471 k-ft). The eccentricity of the point is

defined as

5651.2 k-in37.35 in.

151.3 k

Me

P

[ Point 4 ]

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Select point of straight tension. The maximum tension

in the column is

2

n s y 8 1.27 in 60 ksi

610 k

P A f

[ Point 5 ]


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Point c (in) Pn Mn e

1 - 1548 k 0 0

2 20 1515 k 253 k-ft 2 in

3 17.5 1314 k 351 k-ft 3.2 in

4 12.5 841 k 500 k-ft 7.13 in

5 10.36 585 k 556 k-ft 11.42 in

6 8.0 393 k 531 k-ft 16.20 in

7 6.0 151 k 471 k-ft 37.35 in

8 ~4.5 0 k 395 k-ft infinity

9 0 -610 k 0 k-ft

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In teract ion DiagramColumn Analysis

-1000

-500

0

500

1000

1500

2000

0 100 200 300 400 500 600

M (k-ft)

P(

k)

Use a series of c

values to obtain the

Pn verses Mn.


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Column Analysis

-800

-600

-400

-200

0

200

400

600

800

1000

1200

0 100 200 300 400 500

fMn (k-ft)

fPn

(k)

Max. compression

Max. tension

Cb

Location of the linearlyvarying f.

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ACI Design Aids for Columns

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g

Design Example 8.3

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g p

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Bar splicing in Columns

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p g

Assignment No 1: (Total Marks100 each question carries 50 marks

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Assignment No.1: (Total Marks100 each question carries 50 marks

Design and Rectangular Column to carry dead load of 250K

live load of 350K dead load moment 150ft-K and live loadmoment of 350ft-K Assume material properties.

Determine the main steel required

Determine the ties spacing

Draw final neat to the scale sketch on graph paper

Due Date: Sep,17 2012.

rcc-lec-01

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